2

I want to find out the flaw in this solution to the limit $$\lim_{h\to 0} \frac{\frac{f(a+h)-f(a)}{h}-f'(a)}{h}.$$ The numerator is \begin{align*} & \frac{f(a+h)-f(a)}{h}-f'(a) \\ & =\frac{f(a+h)-f(a+\frac{h}{2})+f(a+\frac{h}{2})-f(a)}{h}-f'(a) \\ & =\frac{1}{2}\left(\frac{f(a+h)-f(a+h/2)}{h/2}+\frac{f(a+h/2)-f(a)}{h/2}\right)-f'(a) \\ & =\frac{1}{2}(f'(a+h/2)+f'(a))-f'(a) \\ & =\frac{1}{2}(f'(a+h/2)-f'(a)). \end{align*} Thus, \begin{align*} \lim_{h\to 0} \frac{\frac{f(a+h)-f(a)}{h}-f'(a)}{h} & = \lim_{h\to 0}\frac{\frac{1}{2}(f'(a+h/2)-f'(a))}{h} \\ & =\lim_{h\to 0}\frac{1}{4}\frac{f'(a+h/2)-f'(a)}{h/2} =\frac{1}{4}f''(a). \end{align*}

Gary
  • 31,845
mingzi xingshi
  • 73
  • 2
    Is it always valid to introduce limit inside limit?

    $$\begin{align} \lim_{h\to0}\frac hh &= 1\ \lim_{h\to0}\frac{\lim_{k\to 0}k}{h} &= \lim_{h\to0}\frac{0}{h} = 0 \end{align}$$

     – peterwhy Jun 08 '22 at 02:35

1 Answers1

2

Why do you think $$\frac{f(a+h)-f(a+h/2)}{h/2}=f'(a+h/2)$$ or $$\frac{f(a+h/2)-f(a)}{h/2} = f'(a)?$$ There's no reason to be like that.

Riemann
  • 1,173